Seeking help with a circuit calculation problem
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All three resistors have a rated power of 10W; R1=10 ohms, R2=40 ohms, and R3=250 ohms. They are connected in series in the circuit. What is the maximum current allowed in this circuit? After thinking about it all night, the answer isn’t 200mA; it’s 5.78mA. I’m not sure how to calculate it Thank you all for your help.
P=I*I*R; R1=1A, R2=0.5A, R3=0.2A. Since they are in series, the maximum current allowed in the circuit is 200mA
I just don’t understand it! Frustrated!
P=I*I*R; I*I=P/R=10000/300=33.33; I=5.77mA
The above formula is the origin of 5.78, which represents the maximum current for a 10W power level. The maximum allowable current for the circuit is as stated in the answer above.
One must adhere to the truth and not deliberately cater to the answers. Verify the conditions and the meaning of the question again.
2nd floor, correct, support! The power in DC resistance is current squared multiplied by resistance; therefore, for R3, it is 10W and 250Ω. Power divided by resistance gives 1/25, and taking the square root yields 1/5, which is obviously 0.2A, or 200mA. Only R3 has the lowest current-carrying capacity; when the three resistors are connected in series, the current flowing through them cannot exceed the value permitted for R3, which is 200mA, of course.
This post was last edited by zhaohh3211 on 2010-10-20 at 14:33. The resistors are connected in series; the current flowing through them cannot exceed the minimum rated current among the three resistors, otherwise one of them will be damaged. Therefore, the rated current of the resistor with the lowest value should be used as the current for the series circuit.
When resistors are connected in series, the current flowing through each resistor is equal; the resistor with a higher value exhibits a greater power dissipation. I3**2=P3/R3=10/250=0.04, I3=0.2A=200mA